Normal push-off zeros are the self-intersection points

The sets. Because $s$ is small, $s(A)$ lies in the interior of the disc bundle, and $\phi$ is injective on the disc bundle. For $p,q\in A$, $\phi (p)=\phi (s(q))$ holds iff $p=s(q)$ as points of the total space (with $p$ viewed as the zero vector over $p$), which forces $p=q$ and $s(p)=0$. Hence $A\cap A_s=\phi (Z(s))$; transversality of $s$ to the zero section is exactly the statement that the vertical differential is surjective at each zero [F1], so the graph $A_s$ meets $A$ transversely there [F2] and by [F4] the tube is a slice chart in which the two submanifolds are the zero section and the graph of the local expression of $s$.

Local model. At $x\in Z(s)$ choose a slice chart of the tube with domain coordinates $(u,v)\in {\mathbb{R}}^a\oplus {\mathbb{R}}^a$ in which $A=\{v=0\}$, $\phi$ is the identity chart, and the orientation of $M$ is the ordered product of the tangent orientation of $A$ and the normal orientation of ${\nu }_A$; this is the tangent-first convention of this pair (An oriented transverse normal bundle orients an embedded submanifold). In these coordinates $A_s$ is the graph of $Ju+O(\mathrm{\mid }u{\mathrm{\mid }}^2)$, where $J={\mathrm{\partial }}_{\nu }{s}_x$ is the vertical derivative. The tangent space of $A_s$ at $x$ is spanned by the vectors $e_j+{\sum }_k{J}_{kj}{e}_{a+k}$; the map $T_{x}A\oplus T_x{A}_s\to T_{x}M$ written in the ordered ambient basis has matrix $\left(\begin{array}{cc}{I}_a& I_a\\ 0& J\end{array}\right)$, whose determinant is $\det J$ (the matrix is square because $2a=n$). By [F3] the sign ${\epsilon }_{(A,A_s)}(\phi (x))$ is the sign of this determinant, proving the displayed identity, including the factor order (first $A$, then $A_s$).

For $a=0$, each point of $A_s=A$ carries the transported point sign ${ϵ}_A$. The local ordered intersection sign is ${ϵ}_A^2{ϵ}_M={ϵ}_M$. The tangent-first normal orientation is ${ϵ}_{\nu }={ϵ}_M{ϵ}_A$, so the vertical map has ray sign ${ϵ}_A{ϵ}_{\nu }={ϵ}_M$ as well. The set and transversality arguments in step 1.1 need no orientations.

Global statement. Adding the finitely many points of $Z(s)$ (finite because $A$ is compact and the zeros of a transverse section are isolated) gives $I(A,A_s)={\sum }_{x\in Z(s)}\mathrm{sign}\det {\mathrm{\partial }}_{\nu }{s}_x$, which is the self-intersection number of The self-intersection number of a complementary-dimensional oriented submanifold. The definition of $A\cdot A$ uses only transverse push-off sections, so no perturbation of a non-transverse section is needed here.